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Question Number 51774 by ajfour last updated on 30/Dec/18

Commented by ajfour last updated on 30/Dec/18

$F i n d m a x i m u m a r e a A i n l i g h t b l u e .$ $(s o u r c e : a j f o u r)$
Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18
$area of OABCDA is=2[(1/2)×h×b+(1/2)×k×a] =hb+ka area if pinnk circle sector=(b^2 /2)×(π−2α) area of yelolow circle sector=(a^2 /2)×(π−2β) 2α+2β=(π/2) α+β=(π/4) area of light blue area hb+ka−[(π/2)(a^2 +b^2 )−(a^2 β+b^2 α)] to my view...tangent from point o to circle radius b...tangent from o to circle radis a should be equal ldngth... so h=k tanα=(b/h) α=tan^(−1) ((b/h)) tanβ=(a/k)=(a/h) β=tan^(−1) ((a/h)) tan(α+β)=((tanα+tanβ)/(1−tanαtanβ))=(((b/h)+(a/h))/(1−((ab)/h^2 )))=(((a+b)/h)/(1−((ab)/h^2 )))=1 ((a+b)/h)=1−((ab)/h^2 ) (a+b)h=h^2 −ab h^2 −h(a+b)−ab=0 h=(((a+b)±(√((a+b)^2 +4ab)))/2) h=(((a+b)+(√((a+b)^2 +4ab)))/2)=L(say) so light blue area hb+ka−[(π/2)(a^2 +b^2 )−(a^2 β+b^2 α)] =L(a+b)−[(π/2)(a^2 +b^2 )−{a^2 tan^(−1) ((a/L))+b^2 tan^(−1) ((b/L))] where L=f(a,b) pls chek upto this step...$
$a r e a o f O A B C D A i s = 2 [\frac{1}{2} \times h \times b + \frac{1}{2} \times k \times a]$ $= h b + k a$ $a r e a i f p i n n k c i r c l e s e c t o r = \frac{b^{2}}{2} \times (π - 2 α)$ $a r e a o f y e l o l o w c i r c l e s e c t o r = \frac{a^{2}}{2} \times (π - 2 β)$ $2 α + 2 β = \frac{π}{2} α + β = \frac{π}{4}$ $a r e a o f l i g h t b l u e a r e a$ $h b + k a - [\frac{π}{2} (a^{2} + b^{2}) - (a^{2} β + b^{2} α)]$ $t o m y v i e w . . . t a n g e n t f r o m p o i n t o t o c i r c l e$ $r a d i u s b . . . t a n g e n t f r o m o t o c i r c l e r a d i s a$ $s h o u l d b e e q u a l l d n g t h . . .$ $s o h = k$ $t a n α = \frac{b}{h} α = {t a n}^{- 1} (\frac{b}{h})$ $t a n β = \frac{a}{k} = \frac{a}{h} β = {t a n}^{- 1} (\frac{a}{h})$ $t a n (α + β) = \frac{t a n α + t a n β}{1 - t a n α t a n β} = \frac{\frac{b}{h} + \frac{a}{h}}{1 - \frac{a b}{h^{2}}} = \frac{\frac{a + b}{h}}{1 - \frac{a b}{h^{2}}} = 1$ $\frac{a + b}{h} = 1 - \frac{a b}{h^{2}}$ $(a + b) h = h^{2} - a b$ $h^{2} - h (a + b) - a b = 0$ $h = \frac{(a + b) \pm \sqrt{{(a + b)}^{2} + 4 a b}}{2}$ $h = \frac{(a + b) + \sqrt{{(a + b)}^{2} + 4 a b}}{2} = L (s a y)$ $s o l i g h t b l u e a r e a$ $h b + k a - [\frac{π}{2} (a^{2} + b^{2}) - (a^{2} β + b^{2} α)]$ $= L (a + b) - [\frac{π}{2} (a^{2} + b^{2}) - {a^{2} {t a n}^{- 1} (\frac{a}{L}) + b^{2} {t a n}^{- 1} (\frac{b}{L})]$ $w h e r e L = f (a, b)$ $p l s c h e k u p t o t h i s s t e p . . .$
Commented by ajfour last updated on 30/Dec/18

$t o u g h t o c h e c k S i r, i h a v e m y s e l f$ $t r i e d a l o n g m r W {S i r}^{'} s l i n e . .!$
	Answered by mr W last updated on 30/Dec/18

	Commented by mr W last updated on 30/Dec/18

	$A_{b l u e} = [a + (a + b) \cos α] [b + (a + b) \sin α] - \frac{{(a + b)}^{2} \sin α \cos α}{2} + \frac{α a^{2}}{2} + \frac{(\frac{π}{2} - α) b^{2}}{2} - \frac{π a^{2}}{2} - \frac{π b^{2}}{2}$ $A_{b l u e} = (a + b) [(a \sin α + b \cos α) + \frac{(a + b) \sin 2 α}{4} + \frac{α (a - b)}{2}] + a b - \frac{π a^{2}}{2} - \frac{π b^{2}}{4} . . . (i i)$ $\frac{{d A}_{b l u e}}{d α} = 0$ $\Rightarrow a \cos α - b \sin α + \frac{(a + b) \cos 2 α}{2} + \frac{a - b}{2} = 0$ $\Rightarrow 2 (a \cos α - b \sin α) + (a + b) \cos 2 α = b - a$ $l e t \frac{b}{a} = μ, t = \tan \frac{α}{2}$ $\Rightarrow \cos α (1 + \cos α) = μ \sin α (1 + \sin α)$ $\Rightarrow μ = \frac{\cos α (1 + \cos α)}{\sin α (1 + \sin α)} = \frac{1 - t}{t (1 + t)}$ $μ t^{2} + (μ + 1) t - 1 = 0$ $t = \tan \frac{α}{2} = \frac{\sqrt{μ^{2} + 6 μ + 1} - μ - 1}{2 μ}$ $\Rightarrow α = 2 \tan^{- 1} \frac{\sqrt{μ^{2} + 6 μ + 1} - μ - 1}{2 μ}$ $\Rightarrow m a x . A_{b l u e} f r o m (i i)$
	Commented by ajfour last updated on 30/Dec/18

	$l e t \tan \frac{α}{2} = t$ $\Rightarrow μ = \frac{(1 - t^{2})}{t {(1 + t)}^{2}} \Rightarrow μ = \frac{1 - t}{t (1 + t)}$ $μ t^{2} + (μ + 1) t - 1 = 0$ $t = \frac{- (μ + 1) + \sqrt{{(μ + 1)}^{2} + 4 μ}}{2 μ}$ $α = {2 \tan}^{- 1} (\frac{- (μ + 1) + \sqrt{{(μ + 1)}^{2} + 4 μ}}{2 μ})$
	Commented by ajfour last updated on 30/Dec/18

	$Y e s S i r, (s o r r y), i t s A b s o l u t e l y r i g h t!$
	Commented by mr W last updated on 30/Dec/18

	$t h a n k y o u s i r! {t h a t}^{'} s c o r r e c t .$
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